Linear simple harmonic oscillator :
The block − spring system is a linear simple harmonic oscillator. All oscillating systems like diving board, violin string have some element of springiness, k (spring constant) and some element of inertia, m.

**Linear
simple harmonic oscillator **

The block − spring system is a linear
simple harmonic oscillator. All oscillating systems like diving board, violin
string have some element of springiness, k (spring constant) and some element
of inertia, m.

**Horizontal
oscillations of spring **

Consider a mass (m) attached to an end
of a spiral spring (which obeys Hooke?s law) whose other end is fixed to a
support as shown in Fig.. The body is placed on a smooth horizontal surface.
Let the body be displaced through a distance x towards right and released. It
will oscillate about its mean position. The restoring force acts in the
opposite direction and is proportional to the displacement.

Restoring force F = −kx

From Newton?s second law, we know that F
= ma

∴ ma = −kx

A =
−kx / m

Comparing with the equation of SHM a =
−ω^{2}x, we get

ω^{2 }= k/m

ω^{ }= rt( k/m )

But T = 2π / ω

Time period T = 2π rt(m/k)

∴ Frequency n =
1/T = 1/2 π rt( k/m )

*Vertical oscillations of a spring*

Fig a shows a light,
elastic spiral spring suspended vertically from a rigid support in a relaxed
position. When a mass ?*m*? is attached
to the spring as in Fig. b, the spring is extended by a small length dl such
that the upward force *F* exerted by
the spring is equal to the weight *mg*.

The restoring force *F *=* k *d*l
*; *k dl *=* mg* ...(1)

where *k*
is spring constant. If we further extend the given spring by a small distance
by applying a small force by our finger, the spring oscillates up and down
about its mean position. Now suppose the body is at a distance y above the
equilibrium position as in Fig. c. The extension of the spring is (*dl* − *y*). The upward force exerted on the body is *k *(*dl *−* y*) and the resultant force*
F *on the body is

*F
*=* k *(*dl - y*)* * * - mg *= -*ky ???..(2)*

The resultant force is
proportional to the displacement of the body from its equilibrium position and
the motion is simple harmonic.

If the total extension produced is (*dl* + *y*)
as in Fig. d the restoring force on the body is *k* (*dl* + *y*) which acts upwards.

So, the increase in the upward force on the
spring is

*k
*(*dl *+* y*)* *−*mg
*=* ky*

Therefore if we
produce an extension downward then the restoring force in the spring increases
by *ky* in the upward direction. As the
force acts in the opposite direction to that of displacement, the restoring
force is − *ky* and the motion is
SHM.

*F
= **-** ky*

* ma = - ky*

a = − k/m y ?(3)

a = −ω^{2}
y (expression for SHM)

Comparing the
above equations, ω = rt(k/m)

But T = 2π / ω =
2π rt(m/k)

From equation
(1) mg = k dl

m/k = dl/g

Therefore time
period T = 2π rt(dl/g) ??(6)

Frequency n =
1/2 π rt(g/dl)

**Case 1 : When two springs are connected in parallel **

Two springs of
spring factors k1 and k2 are suspended from a rigid support as shown in Fig. A
load m is attached to the combination.

Let the load be
pulled downwards through a distance y from its equilibrium position. The
increase in length is y for both the springs but their restoring forces are
different.

If F_{1}
and F_{2} are the restoring forces

F_{1} = −k_{1}y,

F_{2} = −k_{2}y

∴
Total restoring force = (F_{1} + F_{2}) = −(k_{1} + k_{2})
y

So, time period of the body is given by

T = 2π rt(m / k
+k)

If k1 = k2 = k

Then, T = 2π
rt(m/ 2k)

∴
frequency n = 1/2 π . rt(2k/m)

**Case 2 : When two springs are connected in series.**

Two springs are connected in series in two
different ways. This arrangement is shown in Fig. a and b.

In this system
when the combination of two springs is displaced to a distance y, it produces
extension y_{1} and y_{2} in two springs of force constants k_{1}
and k_{2}.

F = −k_{1}
y_{1} ;

F = −k_{2}
y_{2}

where F is the
restoring force.

Total extension,
y = y1 + y2 = −F [1/k_{1} + 1/k_{2}]

We know that F =
−ky

y= - F/k

From the above
equations,

-F/k = −F [1/k_{1} + 1/k_{2}]

Time period = T
= 2π . rt(m(k_{1}+k_{2})/k_{1}k_{2})

frequency n =
1/2 π . rt(k_{1}k_{2 }/ m(k_{1}+k_{2}) )

If both the
springs have the same spring constant,

k1 = k2 = k

n= 1/2 π .
rt(k/2m)

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